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arXiv:1612.02349v1 [math.GR] 7 Dec 2016
On the supersoluble residual
of mutually permutable products
V. S. Monakhov
December 8, 2016
Abstract
We prove that if a group G = AB is the mutually permutable product of the supersoluble subgroups A and B, then the supersoluble residual of G coincides with the
nilpotent residual of the derived subgroup G′ .
Keywords: finite groups, supersoluble subgroup, mutually permutable product.
MSC2010: 20D20, 20E34
All groups in this paper are finite.
If F is a formation and G is a group, then GF is the F-residual of G, i.e., the smallest normal
subgroup of G with quotient in F. A group G = AB is called the mutually permutable product
of subgroups A and B if UB = BU and AV = V A for all U ≤ A and V ≤ B. Such groups
were studied in [1]–[4], see also [5].
We prove the following theorem.
Theorem 1. Let G = AB be the mutually permutable product of the supersoluble
N
subgroups A and B. Then GU = (G′ )N = A, B .
Here N and U are respectively the formation of all nilpotent groups and the formation of
all supersoluble groups, and A, B = h[a, b] | a ∈ A, b ∈ Bi.
We need the following lemmas.
Lemma 1 ([6, 4.8]). Let G = AB be the product of two subgroups A and B. Then
(1) A, B ⊳ G;
(2) if A1 ⊳ A, then A1 A, B ⊳ G;
(3) G′ = A′ B ′ A, B .
1
If X and F are hereditary formations, then, according to [7, p. 337-338], the product
XF = { G ∈ E | GF ∈ X}
is also a hereditary formation.
Lemma 2 ([7, IV.11.7]). Let F and H be formations, G be a group and K ⊳ G. Then
(1) (G/K)F = GF K/K;
(2) GFH = (GH )F ;
(3) if H ⊆ F, then GF ⊆ GH .
If H is a subgroup of a group G, then H G denotes the smallest normal subgroup of G
containing H.
Lemma 3 ([6, 5.31]). Let H be a subnormal subgroup of a group G. If H belongs to a
Fitting class F, then H G ∈ F. In particular,
(1) if H is nilpotent, then H G is also nilpotent;
(2) if H is p-nilpotent, then H G is also p-nilpotent.
Lemma 4. Let G = AB be the product of the supersoluble subgroups A and B. Then
G ≤ A, B .
U
P r o o f. By Lemma 1 (1,3) and Lemma 2 (1),
′
G/ A, B = G′ A, B / A, B = A′ B ′ A, B / A, B =
= A′ A, B / A, B B ′ A, B / A, B .
The subgroups A′ A, B / A, B ≃ A′ / A′ ∩ A, B ,
B ′ A, B / A, B ≃ B ′ / B ′ ∩ A, B
′
are nilpotent [8, VI.9.1] and normal in G/ A, B by Lemma 1 (3), so G/ A, B is nilpotent.
By Lemma 1 (3), A A, B and B A, B are normal in G. In view of the Baer Theorem [9],
G/ A, B is supersoluble. Hence, GU ≤ A, B .
⊠
A Fitting class which is also a formation is called a Fitting formation. The class of all
abelian groups is denoted by A.
Lemma 5 [7, II.2.12]. Let X be a Fitting formation, and let G = AB be the product of
normal subgroups A and B. Then GX = AX B X .
2
P r o o f of Theorem 1. By Lemma 4, GU ≤ A, B . Since U ⊆ NA [8, VI.9.1], by
Lemma 2 (2,3), we have
G(NA) = (GA )N = (G′ )N ≤ GU .
Verify the reverse inclusion. Since
(G/(G′ )N )′ = G′ (G′ )N /(G′ )N = G′ /(G′ )N
is nilpotent,
G/(G′ )N = A(G′ )N /(G′ )N · B(G′ )N /(G′ )N
is supersoluble in view of [1, Theorem 3.8] and GU ≤ (G′ )N . Thus, GU = (G′ )N .
By Lemma 1 (3), G′ = A′ B ′ [A, B] = (A′ )G (B ′ )G [A, B]. The subgroups A′ and B ′ are
subnormal in G by [4, Theorem 1] and nilpotent, therefore (A′ )G (B ′ )G is normal in G and
nilpotent by Lemma 3 (1). In view of Lemma 5 with X = N, we get
GU = (G′ )N = ((A′ )G (B ′ )G )N [A, B]N = [A, B]N .
⊠
Corollary 1.1. Let G = AB be the mutually permutable product of the supersoluble
subgroups A and B. If [A, B is nilpotent, then G is supersoluble.
⊠
The class of all p-nilpotent groups coincides with the product Ep′ Np , where Np is the class
of all p-groups and Ep′ is the class of all p′ -groups. A group G is p-supersoluble if all chief
factors of G having order divisible by the prime p are exactly of order p. The derived subgroup
of a p-supersoluble group is p-nilpotent [8, VI.9.1 (a)]. The class of all p-supersoluble groups
is denoted by pU. It’s clear that Ep′ Np ⊆ pU ⊆ Ep′ Np A.
Theorem 2. Let G = AB be the mutually permutable product of the p-supersoluble
E ′ Np
subgroups A and B. Then GpU = (G′ )Ep′ Np = A, B p .
P r o o f. By Lemma 2,
(G′ )Ep′ Np = (GA )Ep′ Np = GEp′ Np A ≤ GpU .
Verify the reverse inclusion. The quotient group
G/(G′ )Ep′ Np = (A(G′ )Ep′ Np /(G′ )Ep′ Np )(B(G′ )Ep′ Np /(G′ )Ep′ Np )
is the mutually permutable product of the p-supersoluble subgroups A(G′ )Ep′ Np /(G′ )Ep′ Np
and B(G′ )Ep′ Np /(G′ )Ep′ Np . The derived subgroup
(G/(G′ )Ep′ Np )′ = G′ (G′ )Ep′ Np /(G′ )Ep′ Np = G′ /(G′ )Ep′ Np
3
is p-nilpotent. By [4, Corollary 5], G/(G′ )Ep′ Np is p-supersoluble, consequently, GpU ≤ (G′ )Ep′ Np .
Thus, GpU = (G′ )Ep′ Np .
By Lemma 1 (3), G′ = A′ B ′ [A, B] = (A′ )G (B ′ )G [A, B]. The subgroups A′ and B ′ are
subnormal in group G [4, Theorem 1] and p-nilpotent [8, VI.9.1 (a)], hence (A′ )G (B ′ )G normal
in G and p-nilpotent by Lemma 3 (2). In view of Lemma 5 with X = Ep′ Np , we get
GpU = (G′ )Ep′ Np = ((A′ )G (B ′ )G )Ep′ Np )[A, B]Ep′ Np = [A, B]Ep′ Np .
⊠
Corollary 2.1. Let G = AB be the mutually permutable product of the p-supersoluble
subgroups A and B. If [A, B is p-nilpotent, then G is p-supersoluble.
References
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V. S. MONAKHOV
Department of mathematics, Gomel F. Scorina State University, Gomel, BELARUS
E-mail address: [email protected]
4
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